New design of driving l.e.d.s with direct a.c. source

Hello sir

Sir i want to design a circuit with following diagram so as to drive a simple l.e.d. with a house hold a.c. source



Is it okk ?

what changes we have to made ?

are all values ok ?

Values are ok for a led current of about 1 mA may be a little small.
But the Diode in parallel with the led will never conduct so I think it is useless.
What is your purpose in including it ? protection ? balancing current?

albert22 said:

Values are ok for a led current of about 1 mA may be a little small.
But the Diode in parallel with the led will never conduct so I think it is useless.
What is your purpose in including it ? protection ? balancing current?

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yaa protection against the reverse current for l.e.d.s

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sir what should be the minimum value of resistor below 100 k ??????????????????

is there any use of two resistors ?

Considering the power that a resistor can tolerate. You can replace 1 resistor of 4w by two resistors of 2w.
Also esistors have a maximum voltage that can tolerate (besides it power rating) placing several in series avoids exceeding this limitation.
You can choose standard values for resistors from here>
https://www.rfcafe.com/references/electrical/resistor-values.htm

The top diode blocks any reverse voltage anyway so the one across the LED does nothing.

You COULD use one resistor instead of two but it would be a good idea to keep it as it is. The reason is that this circuit gives no isolation from the incoming AC so if you used one resistor, and it was in the neutral side of the AC, the LED would become dangerously 'live' and could kill if you touched it. Using two resistors means the circuit can be connected either way to the AC with less risk of injury. The resistors can be any value you choose. Lower values give more LED current but also more heat, that's why this circuit isn't used commercially. If you imagine a domestic 10W LED lamp drawing say 1.5A, the resistors would be 75 Ohms each and dissipate over 300W of heat!

Brian.

kk so in place of connecting a diode across l.e.d. should i connect the diode in series with low side single resistor in diagram ?

I think you should study Ohms law and learn how current flows in a series circuit. The order of components in the loop makes no difference and all adding a second diode will do is convert a further 0.6V of voltage drop into heat.

You CAN use the reactance of a capacitor to make it more efficient but then you need four diodes and some serious safety issues arise. Your schematic is already potentially lethal and the capacitive method is even worse!

Brian.

nah actually i am talking about diode just because i want to make protection against revese current for l.e.d.
i connected without diode there & so many times it was damaged :x

If you just use resistors with no diode, it WILL be damaged, most LEDS can only withstand about 5V in reverse direction so 230V RMS (about 325V peak) is a bit much for them!
With a single diode in series, no matter where you put it in the chain of components, it should work, the diode will not pass current in the reverse direction except for leakage. If the diode is not a good one, connect a 100K resistor (0.25W will do) across the LED so let it drain away. If you want to keep the resistor solution, it will work better if you use a bridge rectifier with the LED across it's output, you don't need a capacitor as well. This lets the LED light on both positive and negative AC cycles so it will appear brighter and have less flicker.

Whatever you do, please be extremely careful with 230V AC. It is more than enough to kill you and you don't get a second chance.....

Brian.

Your original circuit should work with just one diode in parallel with the led. In this case the negative and positive currents will be almost balanced. But given the low current it this is not necessary. A capacitive solution uses this diode in parallel.

i connected without diode there & so many times it was damaged

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You made me wonder...
The 1N4007 can pass 5uA in reverse and the reverse current of the led is very low too so there is no voltage drop at the resistors. The full peak voltage of the mains (around 300v) is then shared between the diode and the led in series.
This is what kills the led. Definitely the parallel diode is necessary.
I see now that there is an alternative:

If the diode is not a good one, connect a 100K resistor (0.25W will do) across the LED so let it drain away.

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As I said before I think it should work with no diode in series.

Using two resistors means the circuit can be connected either way to the AC with less risk of injury.

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betwixt has a very good point here.

betwixt said:

If you just use resistors with no diode, it WILL be damaged, most LEDS can only withstand about 5V in reverse direction so 230V RMS (about 325V peak) is a bit much for them!
With a single diode in series, no matter where you put it in the chain of components, it should work, the diode will not pass current in the reverse direction except for leakage. If the diode is not a good one, connect a 100K resistor (0.25W will do) across the LED so let it drain away. If you want to keep the resistor solution, it will work better if you use a bridge rectifier with the LED across it's output, you don't need a capacitor as well. This lets the LED light on both positive and negative AC cycles so it will appear brighter and have less flicker.

Whatever you do, please be extremely careful with 230V AC. It is more than enough to kill you and you don't get a second chance.....

Brian.

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ya ya i take always protection against high voltages , thats whu connected so many indicators around the area
so as per your idear diagram is like

**broken link removed**

or something else ?

The drawback to an anti-parallel diode is it doubles the heat dissipation without giving extra brightness. If that's what is desired, a bridge rectifier with the LED across it's output will double the brightness without adding to the heat being produced.

Brian.

betwixt said:

The drawback to an anti-parallel diode is it doubles the heat dissipation without giving extra brightness. If that's what is desired, a bridge rectifier with the LED across it's output will double the brightness without adding to the heat being produced.

Brian.

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so according to you how a circuit diagram it has ? please tell

Simple - a bridge rectifier, rated at 400V or more, the LED wired directly across it's + and - outputs and a resistor in each of the AC inputs to the 230V supply. Dong it that way, the LED get current on both half cycles of the AC and the high reverse voltage problem goes away. You can either use a small bridge rectifier or make one out of four 1N4007 diodes.

It still isn't efficient but the LED is at least twice as bright as before and it's protected against reverse voltage.

Brian.

okk okkk

so is it simple IN 4007 or some same diode with voltage rating 400v ????????????????

If yiu are building the bridge rectifier from four diodes, yes, the 1N4007 is OK Otherwise you can buy a buy a ready made one with a rating of at least 400V.

Brian.

hmm kk sure i ll go for that :smile:

so if want to drive more than a single l.e.d. there say a group of 50 & as per diagram i am connecting two 100k resistor
then how many l.e.d.s i can drive together

a) parallely

b) serially

Firstly - let me warn you again - what you are doing can be VERY dangerous and every part of the circuit is potentially lethal if touched, please make sure it is well insulated.

a) unless you are using specially selected LEDS, never connect them in parallel, they all have very tiny voltage differences and some will hog more current than others.

b) yes, you can do that, add together their voltage drops and work out the resistor values to be (230 - total of all LED voltages)/LED current.

Brian.

okk sir don't worry i take hours on theories & then hours on precations then start my practical in a protected area & i keep a 5 metre distance when powering the assembly , i made a part of my room as like a laoratory with fire extinguishers jet pointing my practical benches don't worry i go for perfections in what i am doing . . thanks for caring

so sir in part b) you are explaning about voltage drop calculations

so how to make it please let me know about the base , what steps i should i ve to follow for final results ???????

I gave you the formula. The resistance in Ohms is the voltage you want to drop divided by the current through the LEDs.

1. Work out the total voltage requirement of the LEDs. The data sheet will tell you "Vf" for the LED. For small LEDs it is usually around 1.6V for 'normal' types and 3.5V for high brightness types. Multiply Vf by the number of LEDs you have wired in series to get the voltage you need. For example five LEDs at 3.5V = 17.5V

2. You are starting with 230V, it is actually 230V RMS but for sake of calculation, just use that number. So the resistors have to drop the difference between 230 and the number you worked out in step 1. Subtract the LED voltage from 230 to find the voltage you need to drop. Using the same example, you want to drop 230 - 17.5 = 212.5V

3. Next work out the resistor needed to drop that voltage at the current you want to use. For small LEDs the current is usually about 5mA, for high brightness LEDs it can be 1A or more. Use the data sheet to find the current for the LEDs you have. The resistor value is Vdrop/ILed, so with the same example and 5mA current the value would be 212.5/0.005 = 42,500 Ohms.

4. You now know the resistor value needed but you are going to use two so each should be half that value, in the example 42,500/2 = 21,250 Ohms. (22K would be the nearest standard value)

5. To pick suitable resistors you need to know how much heat they will dissipate, you can calculate this using I * I * R so for 5mA and 21,250 Ohms they would be 0.53W. For safety, use a resistor rated at about twice that power, in this case 1W rated resistors would be adequate.

Brian.